prove there exists an inverse function

**wilhelm** · December 18th 2012, 08:19 AM

Hello. I'm quite new to calculus and frankly speaking I'm an autodidact. Could you tell me how to solve this:

$f: [\frac{\pi}{2}, \pi) \ni x \rightarrow \frac{1}{sinx} \in R$

Prove that there exists a inverese of $f$ . In what set is $f^{-1}$ differentiable? Calculate $f^{-1'}$

I would really appreciate a thorough explanation. Thank you.

**jakncoke** · December 18th 2012, 10:29 AM

Well, you could easily show that on the interval $[\frac{\pi}{2}, \pi)$ , f is monotonously increasing, since $f'(x)$ exists on the interval, it is continuous on the interval. A continous function which is monotously increasing or decreasing is one to one on that interval and thus also has an inverse. Now if f is differentiable on the interval, $f^{-1}$ is also differentiable given that $f'(x) \not = 0$ at any point on the interval. You can verify that this is true. Thus $f^{-1}$ is differentiable over $[\frac{\pi}{2}, \pi)$

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